HEMATICAL 
CES  LIBRARY. 


UNIVERSITY  OF  CALIFORNIA 
AT  LOS  ANGELES 


685 


Heath's  Mathematical  Monographs 

Issu.-d  under  the  general  ^ .-.••'.• 

Webster  Wells,  S.  B 

Professor  of  Mathematics  ill   Che   Massachuse        i:          .  ':  - 


Some  Noteworthy    :  roperties 

oi  the  Triangu    a  nd 
its  Circle* 


V.   f  i.  BKUX'l   . 


VI .     I 


ii      -    -•  Of  MAIHEMA;      -. 
'-'    STATK  N<  >i  HALS 


D.  C.  Heath  &  Co.,  Publisher,- 


Boston 


New  York 


Chicago 


Number  8; 


Price,   Ten   Cents 


Th 


nner  s 


•.    '  •••  :-.•:-.«  N  •> 


a 


MPmHl^^SSffl^^^^TO^^B^^B 

' 


' 
& 

$9jlniJiflfNm 


^^^^^^^B  K 


Clot) 


d  teachi 


:ructure  oi 


p.l  to  becorru 


little     ; 
ving 
fultoanv 


:'UN 


LIICAGO 


HEATH'S   MATHEMATICAL  MONOGRAPHS 
NUMBER  8 


SOME   NOTEWORTHY   PROPERTIES 

OF  THE  TRIANGLE   AND 

ITS   CIRCLES 


BY 


W.  H.    BRUCE,  A.M.,  PH.D. 

PROFESSOR  OF  MATHEMATICS,   NORTH  TEXAS 

STATE  NORMAL  SCHOOL 

DENTON,  TEXAS 


BOSTON,  U.S.A. 

D.  C.  HEATH  &  CO.,  PUBLISHERS 
1902 


COPYRIGHT,  1902, 
BY  D.  C.  HEATH  &  Co. 


O  /^         "-3'neering  & 
Mathematica/ 
V  X  Sciences 

^ 


THE   TRIANGLE 

1.  The  triangle  is  the  only  rigid  figure ;  that  is 
to  say,  it  is  the  only  polygon  whose  sides  alone 
determine   its    shape    and    size,  or    whose   angles 
determine  the  ratio  of  its  sides  and  of  all  other 
parts. 

2.  Every  triangle  has  three  sides,  three  angles, 
three  altitudes,  three  medians,  three  interior  angle 
bisectors,  three  exterior  angle  bisectors,  and  three 

*  perpendicular  bisectors  of  its  sides.  In  addition 
*  to  these  parts  of  every  triangle,  there  are  certain 
,  other  lines  and  circles  dependent  upon  these  parts ; 
£  and  their  mutual  relations  and  the  variety  of  their 
'  interdependencies  are  fascinating  to  the  student 

just  awakening  to  the  beauties  of  the  science  of 

geometry. 

3.  In  general,  a  triangle  is  determined  by  any 
three   of   its    parts,   one   of   these    parts   being  a 
length,  whether  this  length  be  a  side,  an  altitude, 
the  radius  of  any  one  of  its  numerous  dependent 
circles,  or  any  other  length. 

3 


4  The  Triangle 

4.  One  of  the  simplest  of  the  many  interesting 
relations  between  the  various  parts  of  the  triangle 
is  that  of  the  concurrence  of  certain  corresponding 
lines. 

5.  The  perpendicular  bisectors  of  the  sides  of 
a  triangle  are  concurrent  in  a  point.     This  point, 
being  the   centre  of  the  circumscribed   circle,  is 
called  the  circumcentre  of  the  triangle.     From  the 
definition,  it  is  evident  that  the  circumcentre  of  an 
acute  triangle  is  within   the   triangle ;   that  of  a 
right  triangle  is  the  mid-point  of  the  hypotenuse; 
and  that   of   an    obtuse   triangle  is   without   the 
triangle. 

6.  The  altitudes  of   a  triangle  are   concurrent 
in  a  point  called  the  orthocentre  of  the  triangle. 
Because   the   perpendicular   is    the   shortest   sect 
from  a  point  to  a  given  line,  the  orthocentre  of 
an  acute  triangle  is  within  the  triangle ;  that  of  a 
right  triangle  at  the  vertex  of  the  right  angle ;  and 
that  of  an  obtuse  triangle  without  the  triangle. 

7.  The  medians  of  a  triangle  are  concurrent  in 
that  trisection  point  of  each  which  is  the  second 
from  a  vertex.     This  point  of  concurrence  is  called 
the  centroid  of  the  triangle,  and  it  is  evident  that 
the  point  is  always  within  the  triangle. 

8.  The  bisectors  of  the  interior  angles  of  a  tri- 
angle are  concurrent.     This  point  of  concurrence 


and  its  Circles 


is  the  centre  of  the  inscribed  circle,  and  is  there- 
fore called  the  in-centre  of  the  triangle.  It  is  evi- 
dent that  the  in-centre  is  always  within  the  triangle. 

9.  The  bisectors  of  the  interior  and  the  ex- 
terior angles  of  a  triangle  are  concurrent  four 
times  by  threes. 

\v 


Proof.  Let  ABC  (Fig.  i)  be  any  A,  and  let  the 
bisectors  of  interior  angles  B  and  C  meet  at  O. 
Then  since  BO  is  the  locus  of  points  equidistant 
from  AB  and  BC,  and  CO  the  locus  of  points 
equidistant  from  BC  and  AC,  O  is  equidistant 
from  the  three  sides. 

Similarly  let  bisectors  of  exterior  angles  B  and 
C  meet  at  O',  then  O'  is  equidistant  from  AB,  BC, 
and  AC.  Similarly  for  O"  and  O'". 

10.   Hence  O',  O",  O'",  are  centres  of  circles 
touching  the  sides  of  triangle  ABC.     These  circles 


6  The  Triangle 

are  called  the  escribed  circles  of  the  triangle,  and 
O',  O",  O'",  the  ex-centres  of  the  triangle. 

11.  The  extremities  of  one  side  of  a  triangle 
and  the  feet  of  the  altitudes  to  the  other  two  sides 
are  concyclic  (i.e.  they  lie  on  the  same  circle). 


Fig.  a. 

Proof.  Let  ABC  (Fig.  2)  be  any  A,  and  AD, 
BE,  CPy  its  altitudes.  Then  A,  B,  D,  Et  are  con- 
cyclic,  also  B,  C,  E,  F,  and  A,  C,  D,  F. 

Proof.  The  circle  on  AB  as  diameter  passes 
through  the  vertices  of  the  right  angles  D  and  E. 

Therefore  A,  B,  D,  E,  are  concyclic.  Similarly 
for  B,  C,  E,  F,  and  for  A,  C,  D,  F. 

12.  The  vertex  of  any  angle  of  a  triangle,  the 
feet  of  the  altitudes  to  its  adjacent  sides,  and  the 
orthocentre  are  concyclic. 

Proof.  The  circle  on  BK  (Fig.  3)  as  diameter 
passes  through  vertices  of  right  angles  D  and  F. 

Therefore  B,  D,  K,  F,  are  concyclic.  Similarly 
for  C,  D,  K,  E,  and  for  A,  E,  K,  F. 


Fig.  3. 


13.  Any  two  altitudes  of  a  triangle  cut  each 
other  so   that  the   product   of    the   segments  of 
the  one  is  equal  to  the  product  of  the  segments 
of  the  other. 

Proof.     By  11,  A,  B,  D,  -E(Fig.  2),  are  concyclic. 
.-.  AKxKD  =  BKx  KE. 

(If  two  chords  of  a  O  intersect,  the  product  of  the  segments 
of  the  one  is  equal  to  the  product  of  the  segments  of  the  other.) 

COLLINEARITY. 

14.  Definition.     Collinear  points  are  those  that 
lie  on  the  same  straight  line. 

15.  The    circumcentre,   the    centroid,    and   the 
orthocentre  of  a  triangle  are  collinear,  and  the  sect 
between  the  former  two  is  half  the  sect  between 
the  latter  two. 

Proof.  Let  P  be  the  circumcentre  and  K  the 
orthocentre  of  A  ABC  (Fig.  4).  Draw  MN  and 
PK,  and  also  median  AN  cutting  PK  at  O. 


The  Triangle 


Fig.  4. 

A  MPN  is  similar  to  A  AKC 
(sides  of  one  being  parallel  respectively  to  sides  of  the  other), 


(M  and  N  being  mid-points  of  sides  AB  and  BC). 


ANPO  and  AAOK  are  similar,  and  NP  =  \AK, 
hence,          ON=  \AO,  or  AO  =  \  AN. 

.-.  By  7,       O  is  centroid  of  A  ABC. 
Hence,  the  circumcentre,  P,  the  centroid,  O,  and 
the  orthocentre,  K,  are  collinear,  and  PO  =  %  KO. 

16.  Corollary.      Since  in  similar  triangles  NPM 
and  AKC(Fig.  4),  NM=\AC,  PM=\CK,  we 
may  say, 

The  perpendicular  sect  from  the  circumcentre  to 
a  side  of  the  triangle  is  equal  to  half  the  sect  from 
the  opposite  vertex  to  the  orthocentre. 

17.  The  circle  through   the  mid-points  of  the 
sides  of   a  triangle  passes  also  through  the  feet 


and  its  Circles  9 

of  the  altitudes,  and  bisects  the  sects  between  the 
vertices  and  the  orthocentre. 

Let  L,  My  TV  (Fig.  4)  be  the  mid-points  of  the 
sides,  D,  E,  .Fthe  feet  of  the  altitudes,  of  AAJ3C, 
and  S,  T,  Z  the  mid-points  of  CK,  AK,  and  B K, 
respectively;  then  are  the  nine  points,  Z,  M,  N\ 
D,  E,  F;  S,  T,  Z;  concyclic. 

Proof.  Draw  NT\  APGWand  TGK  are  similar. 
(Sides  of  one  being  parallel  respectively  to  sides  of  the  other.) 

Also,  by  16,  NP  =  \AK=  TK; 

.:  GK  =  GPy  and  GT=  GN. 

Then  since  G  is  the  mid-point  of  PK,  and  T  the 
mid-point  of  AK,  GT=^AP,  radius  of  circum- 
scribed O. 

Hence,  with  G  as  a  centre  and  radius  \AP, 
a  O  will  pass  through  71 

Similarly,  it  is  proved  to  pass  through  5  and  Z. 

But  GN=  GT,  therefore,  this  circle  passes 
through  Ny  and  its  diameter  is  TN. 

Similarly,  it  is  proved  to  pass  through  L  and  M. 

Again,  this  circle  on  77V,  the  hypotenuse  of 
rt.  A  TDN,  as  diameter,  passes  through  D. 

Similarly,  it  is  proved  to  pass  through  E  and  F. 

.'.  L,  M,  N\  D,  E,  F-,  S,  Ty  Z\  are  concyclic. 

18.  The  circle  through  the  nine-points,  men- 
tioned in  17,  is  called  the  nine-point  circle  of  the 
triangle. 


10 


The  Triangle 


19.  Corollary.    The  centre  of  the  nine-point  circle 
of  a  triangle   is  collinear  with  the  circumcentre, 
the  centroid,  and  the  orthocentre,  and  is  midway 
between  the  circumcentre  and  the  orthocentre. 

20.  Corollary.     The    radius    of    the    nine-point 
circle  of  a  triangle  is  half  the  radius  of  the  cir- 
cumscribed circle. 

21.  The   nine-point  circle  of  a  triangle  bisects 
every  sect  between  its  orthocentre  and  its  circum- 
scribed circle. 

C~ 


Fig-  5- 

Let  P  (Fig.  5)  be  the  circumcentre,  and  K  the 
orthocentre  of  A  ABC,  and  HS  a  chord  of  the 
circumscribed  O  through  K,  cutting  the  nine-point 
circle  at  X  and  V.  Then  V  is  the  mid-point  of 
KS,  and  X  the  mid-point  of  KH. 


and  its  Circles  n 

Proof.  From  G,  the  centre  of  the  nine-point 
O,  draw  GX  and  G  V.  Draw  also  PH  and  PS. 

InAPKS,  GV=^PS(2Q). 

InAPKH,  GX  =  \PH=\PS  =  GV. 

Therefore,  no  other  sect  can  be  drawn  from  G 
to  HS  which  will  be  equal  to  GV  or  \PS.  But 
a  sect  from  G  to  mid-point  of  KS  =  \  PS,  there- 
fore, V  is  the  mid-point  of  KS,  and  X  the  mid- 
point of  KH. 

22.  The  sect  from  the  centroid  of  a  triangle  to 
the  circumscribed  circle  is  twice  the  prolongation 
of  that  sect  from  the  centroid  to  the  nine-point 
circle. 

Let  RL  (Fig.  5)  be  a  chord  of  the  circumscribed 
circle  through  Q,  the  centroid  of  A  ABC,  cutting 
the  nine-point  circle  at  E  and  T. 

To  prove  QR  =  2  QT 

and  QL  =  2  QE. 

Proof.     By  15,       PQ  =  \PK. 
By  19,  PG  =  \  PK. 

Hence,    PQ-\  PG,  or  PQ  =  2  GQ. 
In  &PQRand  GQT, 


also  PQ  =  2GQ; 

therefore,  a  sect  from  G,  parallel  to  PR  and  cutting 

QL  is  equal  to  J  PR, 


12 


The  Triangle 


But,  by  20,  GT=\PR  and  GE  =  \PR,  there- 
fore no  other  sect  from  G  to  RL  is  equal  to  \  PR. 

(From  a  given  point  only  two  equal  sects  can  be  drawn  to  a 
given  line.) 

.-.  GT  is  the  parallel  to  PR. 
Hence,  &PQR  and  GQT  are  similar,  and  con- 
sequently QR  =  2QT. 
Similarly,               QL  =  2  QE. 

23.  The  sect  from  the  vertex  of  any  angle  of  a 
triangle  to  the  point  in  which  an  altitude  to  one 
of  its  including  sides  intersects  the  circumscribed 
circle  is  equal  to  the  sect  from  this  vertex  to  the 
orthocentre. 


Fig.  6. 


Let  K  be  the  orthocentre  of  &ABC,  and  let 
AK  prolonged  cut  the  circumscribed  O  at  P,  then 
will  CP=CK. 

Proof.  By  17,  the  nine-point  circle  of  &ABC 
passes  through  D. 

By  21,  KD  =  DP. 


and  its  Circles  13 

.-.  CD  is  the  perpendicular  bisector  of  KP. 
Hence,  CK=CP. 

24.  The  three  circles,  each  of  which  passes 
through  two  vertices  and  the  orthocentre  of  a 
triangle,  are  each  equal  to  the  circumscribed  circle. 
G 


Fig.  7. 

Let  A D,  BE,  CF  (Fig.  7),  be  the  altitudes  and 
K  the  orthocentre  of  A  ABC.  Prolong  CF  to 
meet  the  circumscribed  circle  at  G. 

To  prove  the  circumscribed  circle  of  A.ABK 
is  equal  to  the  circumscribed  circle  of  A  ABC. 

Proof.     Draw  AG  and  B G. 

KG  is  perpendicular  to  AB  by  hypothesis. 

KF=FG  by  21  (F  being  on  the  nine-point  O). 
.',  A  K  =  A  G  and  BK  =  BG. 

Hence,  &AKB  =  &ABG. 

(Having  three  sides  of  one  =  three  sides  of  other.) 


The  Triangle 


/.  The  circumscribed  O  of  A  ,4^  =  the  cir- 
cumscribed O  of  AAJ3G. 

But  the  circumscribed  O  of  A  ABC  is  also  that 
of  A  ABG. 

.'.  The  circumscribed  O  of  AAKB  is  equal  to 
circumscribed  O  of  A  ABC. 

Similarly  for  circumscribed  ©  of  &AKC  and 
BKC. 

25.  Each  centre  of  the  inscribed  or  an  escribed 
circle  of  a  triangle  is  the  orthocentre  of  the  tri- 
angle having  the  other  three  centres  as  vertices. 


Let  O  be  the  in-centre,  O',  O",  O'"  the  ex-centres 
(Fig.  i)  of  &ABC. 

To  prove  O  is  orthocentre  of  A6>'<9"<9'",  O'  the 
orthocentre  of  AOO"O'",  O"  the  orthocentre  of 
AOO'O'",  and  O'"  the  orthocentre  of  A  OO'O". 


and  its  Circles  15 

Proof.  y.BAO"'=y.  CA  O" 

(each  being  half  of  equal  angles). 


(each  being  half  ZBAC). 


Hence,  O'A  is  perpendicular  to  O"O'". 

Similarly,  O"B  and  O'"C  are  altitudes  of 
AO'O"O'". 

/.  O  is  the  orthocentre  of  A  O'O"O'". 

Similarly,  O"C,  O'"B,  and  OA  are  altitudes  of 
AOO"O'",  therefore  O',  their  point  of  concur- 
rence, is  the  orthocentre  of  AOO"O'". 

And  similarly,  O"  is  the  orthocentre  of  &OO'O'", 
and  O'"  is  the  orthocentre  of  A  OO'O". 

26.  The    four    circles    each    of    which    passes 
through   three   of    the   centres   of    the   inscribed 
and  escribed  circles  of  a  triangle  are  equal. 

Let  O,  O',  O",  O'"  (Fig.  i)  be  the  in-centre  and 
the  ex-centres  of  &ABC. 

Then,  by  24  and  25,  the  circle  through  O'OO'" 
is  equal  to  the  circle  through  O'O"O'",  and  simi- 
larly for  the  circles  through  O'OO"  and  O"OO'" 
respectively. 

27.  The    circumscribed   circle   of  a  triangle  is 
the  nine-point  circle  of  each  of  the  four  triangles 
formed  by  joining  the  in-centre  and  the  ex-centres 
of  the  triangle. 


1 6  The  Triangle 

Let  O  (Fig.  i)  be  in-centre,  and  Or,  O",  O'"  the 
ex-centres,  of  A  ABC. 

By  25,  O'A  is±O"O'";  O"B±O'O'";  O'"C±. 
O'O". 

.'.  A,  B,  and  C  are  the  feet  of  the  altitudes  of 
A  OO'O",  OO'O'",  OO"O'",  and  O'O"O"'. 

Hence,  the  circumscribed  O  of  A  ABC  passes 
through  the  feet  of  the  altitudes  of  the  four 
triangles. 

.'.  By  17-18,  the  circumscribed  O  of  A  ABC  is 
the  nine-point  circle  of  the  four  triangles  desig- 
nated. 

28.  The  circumscribed  circle  of  a  triangle  bisects 
the  sects  between  the  in-centre  and  each  ex-centre, 
and  also  the  sect  between  any  two  ex-centres  of 
the  triangle. 

Let  ABC  (Fig.  i)  be  a  triangle,  O  its  in-centre, 
and  O',  O",  and  O'"  its  ex-centres. 

To  prove  the  circumscribed  circle  of  A  ABC 
bisects  OO',  OO",  OO'",  O'O",  O'O1",  and  O"O'". 

Proof.  By  27,  the  circumscribed  O  of  A  ABC 
is  the  nine-point  circle  of  each  of  the  A  OO'O", 
OO'O'",  OO"O"',  and  O'O"O'". 

By  25,  O,  O',  O",  and  O'"  are  each  the  ortho- 
centre  of  the  A  having  the  other  three  as  vertices. 

Therefore,  by  21,  this  nine-point  circle  bisects 
OO',  OO",  OO'",  O'O",  O'O'",  and  O"O'". 


and  its  Circles  17 

29.  The  nine-point  circle  has  become  celebrated 
in  the  history  of  geometry.     Its  discovery  is  usu- 
ally accredited  to  Euler  (1707-1783). 

It  was  first  called  "the  nine-point  circle"  in 
1842  by  M.  Terquem,  editor  of  Nouvelles  Annales, 
this  name  being  derived  from  its  property  of  hav- 
ing as  nine  points  of  its  circumference  the  feet  of 
the  three  altitudes,  the  mid-points  of  the  three 
sides,  and  the  mid-points  of  the  three  sects  from 
the  orthocentre  to  the  vertices. 

In  1822  Professor  K.  W.  Feuerbach  of  Nurem- 
berg published  his  discovery  and  demonstration 
of  the  remarkable  property  of  the  nine-point  circle, 
now  known  as  the  "  Feuerbach  Theorem." 

30.  The    Feuerbach   Theorem.     The  nine-point 
circle  of  a  triangle  is  tangent  to  the  in-circle  and 
each  of  the  ex-circles  of  the  triangle. 

31.  The  nine-point  circle  of  a  triangle  is  tangent 
to  each  of  the  sixteen  inscribed  and  escribed  circles 
of  the  four  triangles  formed   by  joining  the  in- 
centre  and  the  ex-centres  of  the  original  triangle. 

Proof.  By  27,  the  circumscribed  O  of  a  A  is 
the  nine-point  O  of  the  four  A  formed  by  joining 
its  in-centre  and  ex-centres. 

By  the  Feuerbach  Theorem,  the  nine-point  O 
is  tangent  to  each  of  the  inscribed  and  escribed  CD. 


i8 


The  Triangle 


32.  The  feet  of  the  perpendiculars  from  any 
point  on  a  circle  to  the  sides  of  an  inscribed 
triangle  are  collinear. 


Fig.  8. 

Let  ABC  (Fig.  8)  be  an  inscribed  A,  P  any  point 
on  the  circumscribed  O,  and  PD,  PE,  /'/''perpen- 
diculars to  the  sides  of  the  A. 

To  prove  that  Dt  E,  and  Fare  collinear. 

Proof.     Draw  PA,  PBt  PC,  EF,  and  ED. 

In  quadrilateral  PECF,  since  %E  and  ^-F  are 
both  right  A,  the  quadrilateral  is  cyclic  (see  any 
text),  and,  similarly,  PADE  is  cyclic 
(i.e.  its  vertices  are  concyclic). 


(being  inscribed  A  on  the  same  arc  of  O  PECF)  . 


(each  being  supplement  of  %  PCB). 


and  its  Circles  19 


Hence, 

£  PEF+  %  FED  =  £  PAB  + 
But  £  /M£  +  £  />.£•£>  =  st  £ 

(being  opposite  ^  of  quadrilateral  inscribed  in  O  FADE). 

.-.  %  PEF+  %  PED  =  st.  £. 
Hence  DEF  is  a  straight  line. 

33.  Definition.     The  line  joining  the  feet  of  the 
perpendiculars   from  a  point  on  a  circle  to  the 
sides  of  an  inscribed  triangle  is  called  the  pedal 
line  of  the  point  and  the  triangle. 

34.  The  pedal  line  of  a  point  and  a  triangle 
bisects  the  sect  between  such  point  and  the  ortho- 
centre  of  the  triangle. 

Let  PD,  PE,  and  PF  (Fig.  9)  be  J§  to  sides  of 
&ABC  inscribed  in  a  circle  through  P  and  let  O 
be  the  orthocentre  of  A  ABC,  then  will  DF  bisect 
OP. 

Proof.     Prolong  altitude  CG  to  meet  the  O  at  H. 

Draw  AP.  Draw  PH  cutting  AB  at  M  and 
DF  at  N.  Draw  OM. 

Since  £  AFP  and  £  AEP  are  both  rt  £  by  hy- 
pothesis, the  G  on  AP  as  diameter  passes  through 
E  and  F. 

.-.  £  PF.E  =%PAC=%  PHC=  %  FPN 
(PF  and  CH  being  II). 

Hence  A  F/W  is  isosceles,  or  PN  = 


20 


The  Triangle 


H 

^*" 

Fig.  g. 

In  A  OMG  and  HMG, 

OG=  GH 

(as  G  is  on  nine-point  circle  and  H  on  circumscribed  circle 
of  &ASQ. 

.-.  OM=  MH  and  £  OMG  =  £  HMG. 
But   y.HMG  is   complement   of    ~%.MHG  and 
I£  MFN  is  complement  of  £  FPN,  which  equals 


.:  %  HMG  and  its  equals,  ^  OMG  and 

each  =  X  ^fRV. 

/.  F^V=  MN. 

Hence,  P^=  J/7V. 

Also,  since  £  O£f£  =  ^  NFM,  DFis  parallel  to 
OM. 

.-.  Since  DF  bisects  PJ/it  also  bisects  OP. 


and  its  Circles 


21 


35.  The  sect  joining  the  orthocentre  of  an  in- 
scribed triangle  and  a  point  on  the  circumscribed 
circle  and  the  pedal   line  of   the  point  and   the 
circle  intersect  each  other  on  the  nine-point  circle 
of  the  triangle. 

Proof.     By  34,  DF(F\g.  9)  bisects  OP. 
By  21,  the  nine-point  O  of  A  ABC  bisects  OP. 
.'.  the  sect  OP,  the  pedal  line  DF,  and  the  nine- 
point  O  all  have  a  common  point  S. 

36.  The  sect  joining  the   orthocentres  of  two 
triangles  that  have  the  same  base  and  equal  ver- 
tical angles  is  parallel  and  equal  to  the  sect  join- 
ing the  vertices  of  the  vertical  angles. 


Fig.  10. 

In  &ABC  and  DEC  (Fig.  10),  let  4  A  =  %  £>, 
K  the  orthocentre  of  A  ABC,  and  .S  the  ortho- 
centre  of  A  DEC.  Then  will  KS  be  ||  and  =  AD. 


22 


The  Triangle 


Proof.  Since  the  A  have  the  same  base  and  equal 
vertical  angles,  they  have  the  same  circumscribed  O. 

Let  P  be  circumcentre  of  the  triangles,  and 
draw  PMLBC. 


By  16,  AK  and  DS  each  =  2.  PM. 

.'.  AK  =  DSt  hence,  AKSD  is  a  parallelogram. 

.-.  KS  ||  AD  and  =  AD. 

37.  If  ABCD  is  a  cyclic  quadrilateral,  the  sects 
joining  the  orthocentres  of  the  four  triangles 
ABC,  ABD,  ACD,  and  BCD  form  a  quadrilateral 
congruent  with  ABCD. 


Fig.  10. 

Proof.  Let  ABCD  (Fig.  10)  be  a  cyclic  quad- 
rilateral ;  K,  S,  orthocentres  of  A  ABC  and  BDC. 

By  36,  KS  =  and  ||  AD. 

Similarly,  the  sects  joining  any  other  two  ortho- 
centres  =  and  ||  to  a  side  of  ABCD ;  therefore,  the 


and  its  Circles  23 

quadrilateral  whose  vertices  are  the  orthocentres 
of  the  four  triangles  designated  is  equal  to  or  con- 
gruent with  ABCD. 

38.  If  ABCD  is   a   cyclical    quadrilateral,  the 
sects  joining  the  centres  of  the  nine-point  circles 
of  the  triangles  ABC,  ABD,  ACD,  and  BCD  form 
a  quadrilateral  similar  to  ABCD  and  equal  to  one- 
fourth  ABCD. 

Proof.  In  Fig.  10,  draw  PK  and  PS.  By  19, 
the  centre  of  the  nine-point  O  of  A  ABC  is  the  mid- 
point of  PK,  and  the  centre  of  the  nine-point  circle 
of  &.BCD  is  the  mid-point  of  PS. 

The  sect  joining  the  mid-points  of  PK  and 
PS  is  parallel  and  equal  to  \  KS,  and  therefore 
parallel  and  equal  to  \AD. 

Similarly,  the  sects  joining  the  centres  of  the 
other  nine-point  circles  are  respectively  parallel 
and  equal  to  half  a  side  of  ABCD. 

.'.  the  quadrilateral  thus  formed  is  similar  to 
and  equals  \ABCD. 

39.  If  A,  B,  C,  D  are  coney clic,  the  pedal  lines 
of  each  point  and  the  triangle  whose  vertices  are 
the  other  three,  also  the  sects  joining  each  of  these 
points  to  the  orthocentre  of  the  triangle  whose 
vertices  are  the  other  three,  and  the  nine-point 
circles  of  the  four  triangles,  all  pass  through  a 
common  point. 


24  The  Triangle 

Proof.  By  36,  AKSD  (Fig.  10)  is  a  parallelo- 
gram. 

/.  AS  and  DK  bisect  each  other.  Hence,  the 
sect  joining  any  one  of  the  four  points  to  the  ortho- 
centre  of  its  corresponding  triangle  bisects  AS 
and  DK;  therefore  they  all  pass  through  H,  the 
mid-point  of  AS. 

By  21,  the  nine-point  circle  of  each  triangle 
passes  through  the  mid-point  of  the  sect  from  its 
orthocentre  to  the  circumscribed  circle. 

Hence,  the  nine-point  circles  of  the  four  triangles 
all  pass  through  H. 

By  34,  the  pedal  lines  of  each  point  and  the 
triangle  whose  vertices  are  the  other  three  points 
all  pass  through  H,  the  mid-point  of  AS. 

FORMULAS  OF  THE  TRIANGLE 

40.  Let  a,  b,  c,  denote  the  sides  of  a  triangle; 
2  s,  its  perimeter;  A,  its  area;  //,  hn ',  h'",  the 
altitudes ;  m' ,  m" ,  m"',  its  medians  ;  /,  /",  /"',  its 
angle  bisectors  ;  /',  /",  /'",  the  perpendiculars  for 
the  circumcentre  to  the  sides ;  R,  the  radius  of  the 
circumscribed  circle ;  r,  the  radius  of  the  inscribed 
circle ;  r1,  r",  r'",  the  radii  of  the  escribed  circles. 


2      

41.   K  =  - ~vs(s  —  a)  (s  —  £)  (s  —  c). 
a 


42.    A  =  V-r<>  -  «)  0  -  b)  0  -  0- 


and  its  Circles 


ak*      bk"      cti" 
43.    A  =  —  =  --  =  -- 

22  2 


> 

b  +  c 


44. 

45. 

46.  *  = 

4A 

47.  r  =  -• 


-\/bcs(s-a). 


48. 


r  "  = 


__ 

j  —  a  J  —  b  s  —  c 

(For  proofs  of  43-50,  see  any  text.) 


49.    In  an  acute  triangle,  /  +/'  +/"  =  R  +  r. 

Proof. 

C 


Let  ABC  (Fig.  11)  be  a  A,  Z>,  E,  F  mid-points 
of  its  sides,  P  the  circumcentre.  Denote  BC  by  a, 
AC  by  b,  AB  by  c,  PE  by  /,  PF  by  /', 
by  p'». 


26  The  Triangle 

BDPE  is  a  cyclic  quadrilateral. 
Then    BE  x  PD  +  BD  x  PE  =  DE  x 

(The  product  of  the  diagonals  of  an  inscribed  quadrilateral 
is  equal  to  the  sum  of  the  product  of  its  opposite  sides.) 


Or, 

or,  ap'"  +  cp'  =  Z>R.  (i) 

Similarly,  ap"  +  bp'  =  cR,  (2) 

and  bp'"  +  cp"=aR.  (3) 

Again,  ap'  +  bp"  +  cp"  =  2  A  =  ar  +  br  +  cr.      (4) 

Adding  and  factoring, 

(a  +  £  +  ^)(/  +/'  +/")  =  («  +  b  +  ;)(tf  +  r). 

Reducing,      p'  +p"  +/"  =R  +  r. 

50.   In   case  of  an  obtuse  triangle,  p1"   being 
drawn  to  longest  side,  /  +p"  —p"'  =  R  +  r. 

51     I  +  i+-L  =  l. 

W  A  *  .       |  It*  III 

yi         yii         Y  n          <y 

Proof.     By  48, 

1       1        1   _s  —  a  ,s  —  b     s  —  c 
^7  +  /7  +  ^77~     A  A  A 


s 
A 


52        +       +     -  =    - 

'    h'     h"     h"'     r 


, 
Proof. 


and  its  Circles  27 

i+JL       1    =a  +  b+c  _  2j  =  s  _1 
'    h'     k"     h'"          2  A          2  A     A     r 

53  1H_J_..J_  -!+!..  JL 

1    k'*  k"     ft'"      r'     r"^r'" 


Proof.     By51,      +      +       =    . 

Rv««  l  A    l        L«! 

By52,  _7  +  _+___. 

.  !H.J_..J_  -!+!..!_ 

'A'     A"     A'"      /     r"     r'" 

s±C+C-i 

h'*  h"*  h!" 
Proof.          aA'  =  bit"  =  fA'"  =  2  A. 


TT  z.      a         j         " 

Hence,  b  =  -—  and  ^  =  77  77, 

A  "' 

also  a-f>"  +  f" 


Substituting  for  b  and  ^  their  values 


h'fi"      h'p'"       ,, 

or,  P+-jrr  +  it!*-=h' 

P'     P"     P'" 
Hence,  F  +  F"  +  F7"  = 

55.   rr'  +  fr"  +  rr>"  +  r'r"  +  r'r"'  +  r"r"' 
—  ab-\-ac-\-  be. 


28  The  Triangle 

Substituting  for  r,  r',  r",  r'",  their  values  in  49 
and  48,  the  first  member  of  the  equation  becomes 

A2  A2  A2  A2 

s(s  —  a)     s(s  —  b)     s(s  —  c)     (s  —  a)(s  —  b) 

A2  A2 

(s  —  a)(s  —  c)     (s  —  &)(s  —  c) 

Substituting  value  of  A  in  42,  and  reducing, 

=  (s-  b}(s  -c)  +  (s-  a)(s  -  c)  +  (s  -  a)(s  -  b) 

+  s(s  —  c)  +  s(s  —  b}  +  s(s  —  a) 
=  (s  -  *)[(•*  -  <0  +  0  -  ay]  +  (s  -  a}[(s  -  c}  +  s] 

+  sl(s-c}  +  (s-b}-] 
=  (s-b%2s-a-  c*)+(s  -  ay(2s-  c}+s(2s-b-c} 

=  bs  —  b2  +  as  +  bs  —  a2  —  ab  +  as 


=  ab  +  ac  +  be 


56.  A=V?rW". 


DESCRIPTIVE  PRICE  LIST 
OF  WELLS'  NEWEST  BOOKS 

BOUND  IN  HALF  LEATHER. 
LARGE  PAGE.     CLEAR  TYPE. 

ACADEMIC  ARITHMETIC 

348  pages.     Price,  $1.00. 

ESSENTIALS  OF  ALGEBRA 

With  or  without  answers.  3,300  problems.  375  pages. 
Price,  #1.10. 

NEW  HIGHER  ALGEBRA 

With  or  without  answers.  460  pages.  Price,  $1.32. 
Pages  1-358  identical  with  the  Essentials  of  Algebra. 
The  latter  part  treats  adranced  topics. 

COLLEGE  ALGEBRA 

With  or  without  answers.  583  pages.  Pri«e,  $1.50. 
PART  II,  beginning  with  Quadratics,  341  pages. 
Price,  $i. 32. 

ESSENTIALS  OF  GEOMETRY 

Plane  and  Solid.  402  pages.  Price,  $1.25.  Bound 
separately,  75  cents  each. 

COMPLETE  TRIGONOMETRY  (1900) 

148  pages.  Price,  90  cents.  With  Four-Place  Tables, 
$1.08. 

NEW  PLANE  TRIGONOMETRY  (1900) 

ioo  pages.  Price,  60  cents.  With  Four-Place  Tables, 
75  cents. 

NEW  SIX-PLACE  LOGARITHMIC  TABLES 

Cloth,    go  pages.    Price,  60  cents. 

NEW  FOUR-PLACE  TABLES 

Cloth.    26  pages.     Price,  25  cents. 

D.   C.    HEATH   &   COMPANY 

BOSTON  NEW  YORK  CHICAGO          LONDON 


Essentials  of  Geometry 

(Plane  and  Solid} 
BY  WEBSTER  WELLS 

Massachusetts  Institute  of  Technology 


r  I  "»HE  ESSENTIALS  OF  GEOMETRY  offers  a 
A  practical  combination  of  more  desirable 
qualities  than  any  other  geometry  ever  pub- 
lished. It  excels  in  exact  and  concise  defini- 
tions;  in  arrangement  and  grading ;  in  brevity 
and  uniformity  in  the  statement  of  theorems ; 
in  helpful  and  suggestive  figures ;  in  variety 
and  abundance  of  original  exercises,  with  fig- 
ures and  hints  wherever  needed ;  in  method 
of  proof  characterized  by  simplicity,  brevity, 
and  originality.  Self-reliance,  resourcefulness, 
and  ingenuity  are  the  results  of  its  use. 


Half  Leather.    399  pages.     Introduction  price,  $1.25 
Bound  separately:   Plane,  $0.75.    Solid,  $0.75 


D.  C.   HEATH   &  COMPANY 

Boston  New  York  Chicago  London 


THE  VITAL  DIFFERENCE 

lies  in  this: 


Wells's  Essentials  of  Plane  and  Solid 
Geometry  is  characterized  by  the  in- 
ductive method  of  demonstration,  and 
always  requires  the  student  to  do  for 
himself  the  maximum  amount  of  rea- 
soning and  thinking  of  which  he  is 
capable,  while  most  geometries  require 
of  the  pupil  in  his  demonstrations  little 
personal  power  except  that  of  memo- 
rizing. They  give  the  reasons  for 
statements  ;  while  Wells,  whenever 
the  student  should  be  able  to  give  the 
reason  for  himself,  always  asks  why 
the  statement  is  true.  No  other  geom- 
etry develops  so  strongly  the  power  of 
vigorous  independent  thought. 


D.  C.  HEATH  &  COMPANY 

Boston  New  York  Chicago  London 


Essentials  of  Algebra 

By  WEBSTER  WELLS 

Massachusetts  Institute  of  Technology 


THE  ESSENTIALS  OF  ALGEBRA  is  the  sim- 
plest, most  logical,  most  philosophical, 
and  most  carefully  graded  algebra  now 
available  for  American  schools.  The  method 
of  presenting  the  fundamental  topics  differs  at 
several  points  from  that  usually  followed,  and 
particular  care  has  been  taken  to  give  a  com- 
prehensive and  intelligible  development  of  the 
Theory  of  Algebra. 

The  examples  and  problems  number  over  3300 
and  all  of  them  are  new.  They  are  especially 
numerous  in  the  important  chapters  on  factoring, 
fractions  and  radicals. 

In  the  accuracy  of  definitions,  clearness  of 
demonstrations,  number  and  grading  of  problems, 
and  logical  arrangement,  this  algebra  is  une- 
qualled. It  is  sufficient  in  scope  to  meet  fully  the 
most  rigid  requirements  now  made  in  secondary 
schools. 


Half  leather.    375  pages.    Introduction  price,  $1.10. 

D.  C.  HEATH  &  COMPANY 

Boston  New  York  Chicago  London 

M 


^d&s*'3tfi!StB<i 


Mathemati 


1C 


SffifebSafiSf 

arton's  Theory  o 
Bowser's  Academic  Algebra,    for  s 

leg     '        ra.      i 
•'s  Plane  and  Solid  Geome: 

i  Elemon  [31          !  Sp 

Bowser's  Treatise  on  Plane  and  Sp 


ithmic  Tables 


Lefev 

Lyma 
McCu 


Nichols's  CalCuiU.-..      !;.*>.,!  til  tiai^.uf!  I  u; 

Osborne's  Pifferential  and  Integral  Caku 
Peterson  and  Baldwin's  Problems  in  Al^ 

i^^sW^^BW''w^5B^H^^sWWli*^l 
Scbwatt's  Geometrical  Treatment  of  Cur 


•-    lls's  Academic  Algebra. 

.•  -'  i. 


's  College  Algebra.    .  51.50.    Pan  H,  ixi^'iti! 

?^ 

W  ells  s)     '  '.juts  otGevmn   .      ,          .     ,    - 
,,"_•  sTU     -   "•     •    u  a    •  ph         il  Trigom 

$i.c(o.     With  six  place  tabiies,,  $i'.aj.     "With  Ry! 


Wells's  New  Six-Place  Lojjarit'r.-       i  ibles 
Bella's  Pour--  :        .     : 


D.G.  HEATH&  CO.,  Publishers,  Be  toi 


EMAT1CAL  MONOGRAPHS 


SURD  T7KDER  THS  GENERAL  EDITORSHIP  OF 

WEBSTER  WELLS,  SB. 

.••  ••      •;.,.'.       -        ;    i'.la.sxach'ui.elt    .'    stitutf  ot    ' 


achi  rs    oi     mathi  ,'...-.'     .,    by  presenting 

'   •  :.-   .  -    upon    the   hisi 
^^f§P^fi^P|ffi^ffiH8lmHS       •    •    l      :      ''  ;'- 


GEOMETRICAL  THEOREMS  AND  PROBLEMS    AND 
THEIR  HISTORY. 


1 

inr.g  the  Circle 


a.    FAMOUS  GEOMETRICAL  THEOREM 

"  in  Propo:      i     '  •    .     •, 

FAMOUS  GEOMETRICAL  THEOREMS.     B 


-'AMOUS  GEOMETRICAL  THEOREMS.  '    ;     •  W.K   •;    ; 

:  n  of   i  .         ii        [pthcmati  :al    In<   npti,  n     ipou   ;>: 

Tombstone  of  Liidoiph  Van  Ceulcn. 

oar   TEACHING  GEOMET;:  ,         ;    MILNEK. 

C  •-  '•'•  ,  ES  .       :     Professot  1     J       •  :    .    :  d;a    a  (  'nivi  rsitjr. 
FACTORING.     Ey  Frofes:,-.  >r.-  VVEBSTi      iVEJU    . 

PRICE,  io  CENTS  EACH 


D.  C,  HEATH  &  CO.,  Publishers,  Boston,  New  Yorfc, 


A     000  427  771     1 


DEC  1 8  1 

SEP   i 
SEP  2  . 


SEP  24 
iovio 

NOV  1 0 

DEC  2   1954 


APR 


958 


«Af 


14 


HB 


AIOV2 


NOV  4  -1959 


I95J 


lOw-4,'28 


